792: For Group and Finite-Order Element, Conjugate of Element Has Order of Element

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description/proof of that for group and finite-order element, conjugate of element has order of element

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Topics

About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of order of element of group.
• The reader admits the proposition that for any group and its any finite-order element, the order power of the element is $1$ and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.
• The reader admits the proposition that for any group and any element, if there is a positive natural number to power of which the element is 1 and there is no smaller such, the subgroup generated by the element consists of the element to the non-negative powers smaller than the number.

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Target Context

• The reader will have a description and a proof of the proposition that for any group and any finite-order element, any conjugate of the element has the order of the element.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$G$: $\in \{\text{ the groups }\}$
$g$: $\in G$, such that $|(g)|<\mathrm{\infty }$
$g^{\prime }$: $\in G$
//

Statements:
$|(g^{\prime }g{g}^{\prime -1})|=|(g)|$
//

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2: Natural Language Description

For any group, $G$, any finite-order element, $g\in G$, and any element, $g^{\prime }\in G$, $|(g^{\prime }g{g}^{\prime -1})|=|(g)|$.

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3: Proof

Whole Strategy: Step 1: let $n:=|(g)|$ and see that $(g^{\prime }g{g}^{\prime -1}{)}^n=1$; Step 2: see that there is no $k\in \mathbb{N}\setminus \{0\}$ such that $k<n$ and $(g^{\prime }g{g}^{\prime -1}{)}^k=1$; Step 3: conclude the proposition.

Step 1:

Let $n:=|(g)|$.

$g^n=1$, by the proposition that for any group and its any finite-order element, the order power of the element is $1$ and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.

$(g^{\prime }g{g}^{\prime -1}{)}^n=g^{\prime }g{g}^{\prime -1}{g}^{\prime }g{g}^{\prime -1}...g^{\prime }g{g}^{\prime -1}=g^{\prime }g1g1g...1g{g}^{\prime -1}=g^{\prime }{g}^n{g}^{\prime -1}=g^{\prime }1g^{\prime -1}=g^{\prime }{g}^{\prime -1}=1$.

Step 2:

Let us see that there is no $k\in \mathbb{N}\setminus \{0\}$ such that $k<n$ and $(g^{\prime }g{g}^{\prime -1}{)}^k=1$.

Let us suppose that there was such a $k$.

$g^k=g^{\prime -1}{g}^{\prime }{g}^k{g}^{\prime -1}{g}^{\prime }=g^{\prime -1}(g^{\prime }g{g}^{\prime -1}{)}^k{g}^{\prime }=g^{\prime -1}1g^{\prime }=g^{\prime -1}{g}^{\prime }=1$, a contradiction against $|(g)|=n$, by the proposition that for any group and its any finite-order element, the order power of the element is $1$ and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.

Step 3:

$(g^{\prime }g{g}^{\prime -1})=\{1,g^{\prime }g{g}^{\prime -1},...,(g^{\prime }g{g}^{\prime -1}{)}^{n-1}\}$, by the proposition that for any group and any element, if there is a positive natural number to power of which the element is 1 and there is no smaller such, the subgroup generated by the element consists of the element to the non-negative powers smaller than the number.

That means that $|(g^{\prime }g{g}^{\prime -1})|=n=|(g)|$.

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References

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